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Thermo solution8 - Copy - 650:351 Thermodynamics Doyle...

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650:351 Thermodynamics Doyle Knight Department of Mechanical and Aerospace Engineering Rutgers - The State University of New Jersey New Brunswick, NJ USA Tel: 732 445 4464 · Email: [email protected] Office hours: Tuesday and Friday, 9:00 am to 11:30 am and by appointment
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Homework Assignment No. 8 Problem 9.35 A flow of 2 kg/s saturated vapor R-22 at 500 kPa is heated at constant pressure to 60 C. The heat is supplied by a heat pump that receives heat from the ambient at 300 K and work input. Assume the process is reversible and find the rate of work input. Solution The control volume is the heat exchanger. From the conservation of mass, ˙ m i = ˙ m e = ˙ m . From the First Law of Thermodynamics, 0 = ˙ Q H + ˙ mh i ˙ mh e since the change in kinetic and potential energy across the heat exchanger is negligible. From Table B.4.1, for saturated R-22 vapor, h g = 249 . 95 kJ/kg at p = 497 . 6 kPa and h g = 251 . 73 kJ/kg at p = 583 . 8 kPa 650:351 Thermodynamics Page 1
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Homework Assignment No. 8 Problem 9.35 Thus, interpolating to p = 500 kPa, h 1 = 250 . 0 kJ/kg and s 1 = 0 . 9267 kJ/kg-K Note that T 1 = 0 . 14 C. Since T 2 = 60 C and p 2 = p 1 , the R-22 exits the heat exchanger as superheated vapor. From Table B.4.2, at p = 500 kPa and T = 60 C, h 2 = 293 . 22 kJ/kg and s 2 = 1 . 0696 kJ/kg-K Thus, ˙ Q H = ˙ m ( h 2 h 1 ) = (2)(293 . 22 250 . 0) = 86 . 44 kJ/s Consider a control volume enclosing the heat pump and heat exchanger. The First Law of Thermodynamics is 0 = ˙ Q L ˙ W + ˙ mh i ˙ mh e Why is ˙ Q H not included ? The Second Law of Thermodynamics is 0 = ˙ m i s i ˙ m e s e + ˙ Q L T L + ˙ S gen Why is ˙ Q H not included ? 650:351 Thermodynamics Page 2
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Homework Assignment No. 8 Problem 9.35 Since the problem statement indicates that the process is reversible, ˙ S gen = 0. Thus, 0 = ˙ ms i ˙ ms e + ˙ Q L T L Thus, ˙ Q L = ˙ mT L ( s 2 s 1 ) = (2)(300)(1 . 0696 0 . 9267) = 85 . 74 kW From the First Law of Thermodynamics applied to the control volume enclosing the heat pump and heat exchanger, ˙ W = ˙ Q L + ˙ m ( h i h e ) = 85 . 74 + (2)(250 293 . 22) = 0 . 7 kW Note that work is done on the control volume and hence ˙ W is negative. 650:351 Thermodynamics Page 3
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Homework Assignment No. 8 Problem 9.36 A reversible steady-state device receives a flow of 1 kg/s air at 400 K and 450 kPa, and the air leaves at 600 K and 100 kPa. Heat transfer of 800 kW is added from a 1000 K reservoir, 100 kW is rejected at 350 K, and some heat transfer takes place at 500 K. Find the heat transferred at 500 K and the rate of work produced. Solution The control volume is the device. The conservation of mass is ˙ m i = ˙ m e = ˙ m The First Law of Thermodynamics is 0 = ˙ Q 1 ˙ Q 2 + ˙ Q 3 ˙ W + ˙ mh i ˙ mh e Note sign for ˙ Q 2 which is assumed positive for heat removed The Second Law of Thermodynamics is 0 = ˙ ms i ˙ ms e + summationdisplay ˙ Q T + ˙ S gen 650:351 Thermodynamics Page 4
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Homework Assignment No. 8 Problem 9.36 Since the problem statement indicates that the process is reversible, ˙ S gen = 0. Thus, 0 = ˙ ms i ˙ ms e + ˙ Q 1 T 1 ˙ Q 2 T 2 + ˙ Q 3 T 3 where T 1 = 1000 K, T 2 = 350 K and T 3 = 500 K. Also, ˙ Q 1 = 800 kW and ˙ Q 2 = 100 kW. Thus, ˙ Q 3 = T 3 ˙ m ( s e s
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